If $a \cos A = b \cos B$,then $\Delta ABC$ is

The two adjacent sides of a cyclic quadrilateral are $2$ and $5$ and the angle between them is $60^{\circ}$. If the area of the quadrilateral is $4\sqrt{3}$,then the perimeter of the quadrilateral is

$ABCD$ is a trapezium such that $AB$ and $CD$ are parallel and $BC \perp CD$. If $\angle ADB = \theta$,$BC = p$ and $CD = q$,then $AB$ is equal to:

In triangle $ABC$,$\frac{\tan A}{2} = \frac{\tan B}{3} = \frac{\tan C}{4}$,then the value of $\sec^2 A + \sec^2 B + \sec^2 C$ is:

Observe the following statements:
$(I)$ In $\triangle ABC$,$b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} = s$
$(II)$ In $\triangle ABC$,$\cot \frac{A}{2} = \frac{b+c}{a} \implies B = 90^{\circ}$
Which of the following is correct?

In $\triangle ABC$,with usual notations,if $a \cos B = b \cos A$ and $a \cos C \neq c \cos A$,then the area of $\triangle ABC$ is . . . . . . sq. units.